2nd law of thermodynamics for non-quasistatic processes

Question

The 2n law of thermodynamics can be stated in terms of entropy as follows

$dS \geq \frac{dQ}{T},$

which holds for all quasistatic processes (reversible and irreversible ones).

Is there a generalization of this statement to a general process between two equilibrium states $e_1$ and $e_2$ (a non-quasistatic process)? I.e. can one write down a similar inequality for $\Delta S = S(e_2) - S(e_1)$ (linking it to $\Delta Q$ and so on)? Or at the very least, is it possible to derive the well-known $\Delta S \geq 0$ for an isolated system?

I'm aware of the fact that one can always write $\Delta S = \int_{\gamma} \frac{dQ}{T}$ for any reversible process $\gamma$ driving the system from to $e_1$ to $e_2$. However, it's not obvious how to exploit this, if at all.

Answer 1

The Clausius inequality holds not only for quasi static process but for all processes carried out on a closed system. That is, it holds if it is expressed correctly: $$\Delta S\geq\int{\frac{dQ}{T_I}}$$ where $T_I$ represents the temperature at the interface between the system and its surroundings (where the heat transfer dQ occurs). The integral in the equation represents physically the entropy transferred into the system from the surroundings. The reason that $\Delta S$ is greater than this for an irreversible process is that, in addition to the entropy transferred in from the surroundings, entropy is also generated within the system. Such entropy generation does not occur in a system experiencing a reversible process.